In my textbook "Elementary Number Theory with Applications" by Thomas Koshy on pg. 16, there is an example given just after the well ordering principle:
Prove that there is no positive integer between $0$ and $1$.
My question is how can you even start this proof? I checked the book and google for a formal definition of integers, but they are pretty vague. The book just says $\{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$, and some of the websites say that it's a number without a fractional component. Then what is a fractional component? I took a course on Algebra and we defined them as equivalence classes... but to obtain that, it would require the knowledge of basic things like $1$ is the least positive integer. So I'm getting stuck in this loop. Here's what the textbook says.
Proof. (as in textbook)
Suppose there is a positive integer a between $0$ and $1$. Let $S = \{n \in \mathbb{Z}^+ | 0 < n < 1\}$. Since $0 < a < 1$, $a \in S$, so $S$ is nonempty. Therefore, by the well-ordering principle, $S$ has a least element $l$, where $0 < l < 1$. Then $0 < l^2 < l$, so $l^2 \in S$. But $l^2$ < $l$, which contradicts our assumption that $l$ is a least element of $S$. Thus, there are no positive integers between $0$ and $1$.
However, we have no definition of $l^2$ (and we can't say anything about its order).
I'm just simply not getting convinced by this proof. Can someone help?
I think this is a weird example chosen to illustrate the version of induction that uses the well ordering of the positive integers. That means you can assume whatever you need about arithmetic, including squaring, and the fact that for real numbers $a < 1$ implies $a^2 < a$.
Then, as the argument says, the square of some (hypothetical) integer between $0$ and $1$ will be less than $a$.